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For a rapidly rotating neutron star, if consider the star as a sphere, redshift at the equator surface will be larger than at the pole. But if consider the star as an obsolate sphereiod (the ellipsoid varies according to the rotation velocity), then how about the redshift at equator compare to the pole surface? Whether the obsolate decrease the redshift at equator? Similar is the effective gravity, if consider the obsolate effect, how about the gravity difference between the equator and the pole compare to that a sphere?

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  • $\begingroup$ Why would the redshift at the equator be larger than at the pole for a spherical star? $\endgroup$
    – ProfRob
    Dec 4 '20 at 16:51
  • $\begingroup$ Imagine a rotating sphere. Its equator partially blueshifted, partially redshifted, depending on the angle. Its pole is not shifted in any direction (by doppler, there is still a gravitational redshift due to its gravity). How this differs if the star is an oblate spheroid? I think, it does not. Please extend your question to make it more clear. $\endgroup$
    – peterh
    Dec 4 '20 at 18:38
  • $\begingroup$ But @peterh-ReinstateMonica that is a variable Doppler shift, not a redshift. $\endgroup$
    – ProfRob
    Dec 5 '20 at 0:26
  • $\begingroup$ From this paper: arxiv.org/pdf/1310.0987.pdf you can see that the redshift is latitude dependent for a rotating neutron star. $\endgroup$
    – Chen
    Dec 5 '20 at 9:34
  • $\begingroup$ BTW, not all fast spinning neutron stars are necessarily oblate spheroids. There is evidence to suggest that the extremely intense magnetic fields of magnetars can squeeze them into prolate spheroids, as I mentioned here. $\endgroup$
    – PM 2Ring
    Dec 5 '20 at 15:17
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The redshift factor is given in (Asaoka & Hoshi 1987) and depends on latitude and radial coordinates. The equilibrium surface may however be an ellipsoidal surface of constant redshift and frame dragging (Marsh 2014).

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  • $\begingroup$ Thank you @Anders Sandberg, Marsh 2014 is the article i'm looking for. If we calculate neutron star redshift z0 as a sphere, then z_equator should vary between the z0_equator and z0_pole, according to the obsolateness. Then is that the boundary of rotation? And how about the effective gravity g at the star surface when it become obsolate? At the constant redshift spheroid surface, g is also constant? $\endgroup$
    – Chen
    Dec 5 '20 at 9:32
  • $\begingroup$ I don't understand: how can you have oblateness on a sphere? $\endgroup$ Dec 5 '20 at 10:48
  • $\begingroup$ What i said has something wrong: z0 is got when the star is sphere, both z0_equator and z0_pole. Then is that the boundary of rotation, i.e after the equator redshift equals that of the pole, can the star rotate more faster and make redshift continue to decrease, smaller than at the pole? And how about the effective gravity g at the star surface when it become obsolate? At the constant redshift spheroid surface, g is also constant? $\endgroup$
    – Chen
    Dec 5 '20 at 13:51

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